Probability Question Help?

If i can choose five characters at random from {A, B, C, D, E, F} to form a string what is the probability this string has an odd number of vowels? How would I work this out? I assume its something to do with P(n,r) and C(n,r). I dont know the difference between these though!





Any help would be very much appreicated!

Probability Question Help?
P(n,r) = permutations - order makes a difference


C(n.r) = combinations - order does not make a difference.





I'm going to make this easy and assume that once you pick a letter, it can not be picked again. Also since you only care about the number of vowels and not their order in picking, then order does not matter.





C(6,5) = 6!/(5!*1!) = 6





There are only six possible outcomes (ignoring order):





ABCDE


ABCDF


ABCEF


ABDEF


ACDEF


BCDEF





There are only 2 outcomes were you have either the A or the E, not both.





So Probability (odd # of vowels) = 2/6 = 1/3 = 33%





Now, if you can pick the same letters over and over again, that makes the problem much harder. The total number of possible solutions would be 6^5 = 7776 five letter strings.





The odds of getting a vowel = 1/3


The odds of getting a consanant = 2/3





Prob (1 vowel and 4 cons) = 5C1*(1/3)^1*(2/3)^4 = 5!/(4!*1!)*(1/3)*(2/3)^4 = 5*(1/3)*(16/81) = 0.329





Prob (3 vowel and 2 cons) = 5C3*(1/3)^3*(2/3)^2 = 5!/(2!*3!)*(1/3)^3*(2/3)^2 = 10*(1/27)*(4/9) = 0.165





Prob (5 vowel and 0 cons) = 5C5*(1/3)^5*(2/3)^0 = 5!/(5!*0!)*(1/3)^5*(2/3)^0 = 1*(1/243)*1 = 0.004





P(odd # vowels) = 0.329 + 0.165 + 0.004 = 0.498 = 49.8%
Reply:nCr = n!/(r!(n-r)! is the number of ways of choosing r things out of a field of n things.


nPr = n!/(n-r)! is the number of arrangements of r things from a field of n things.





the answer depends upon whether repeated letters are allowed.





assume not.


then there are 6C5 =6 number of ways of choosing 5 letters out of the six given.


consider the omitted letter. 1/3 of the time it is a vowel and the string has 1 (an odd number) of vowels. 2/3 of the time, it is a consonant, and the string has an even number (2) of vowels.





assume repeated letters are allowed.


each character has a 1/3 chance of being a vowel.


choose the first char. 1/3 odd, 2/3 even


choose the second char.


1/3*1/3 + 2/3*2/3 even (both vowels or both consonants)


4/9 odd, 5/9 even


choose the third


4/9*1/3 + 5/9*2/3 even


14/27 even, 13/27 odd


choose the 4th


14/27*1/3 + 13/27*2/3 odd


40/81 odd, 41/81 even


choose the 5th


40/81*1/3 + 41/81 *2/3 even


122/243 odd, 121/243 even





just over 1/2 odd, just under 1/2 even